MATH PROBLEMS, WITH SOLUTIONS

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1 MATH PROBLEMS, WITH SOLUTIONS OVIDIU MUNTEANU These are free online notes that I wrote to assist students that wish to test their math skills with some problems that go beyond the usual curriculum. These notes can be used as complimentary to an advanced calculus or algebra course, as training for math competitions or simply as a collection of challenging math problems. Many of these are my own creation, some from when I was a student and some from more recent times. The problems come with solutions, which I tried to make both detailed and instructive. These solutions are by no means the shortest, it may be possible that some problems admit shorter proofs by using more advanced techniques. So, in most cases, priority has been given to presenting a solution that is accessible to a student having minimum knowledge of the material. If you see a simpler, better solution and would like me to know it, I would be happy to learn about it. Of course, I will appreciate any comments you may have. I have included problems from linear algebra, group theory and analysis, which are numbered independently. In the future, I plan to expand this set and include problems from more fields as well.

2 2 OVIDIU MUNTEANU. Algebra. For two n n invertible matrices A, B such that AB + BA = O, show that I, A, B and AB are linearly independent. Solution: We can proceed directly: we want to show that if a, b, c, d are such that ai + ba + cb + dab = O, then a = b = c = d =. Denote by C = AB, and observe that since AB + BA = O, we get AC + CA = O and BC + CB = O as well. Now we multiply ai + ba + cb + dab = O to the left by A and get which can be written as As A is invertible, this implies aa + ba 2 + cab + dac = O, (ai + ba cb dc)a = O. I + ba cb dc = O. Together with ai + ba + cb + dc = O, it yields { ai + ba = O cb + dc = O. The last equation can be simplified again, using that B is invertible. Indeed, implies O = cb + dc = (ci + da)b { ai + ba = O ci + da = O. Each of these equations imply the coefficients are zero. Indeed, if there exist u and v so that ui + va = O and if v, then A = λi, for λ = u. But then v O = AB + BA = 2λB, therefore λ =, which is a contradiction. 2. Let A, B be two n n matrices that commute. For any eigenvalue α C of A + B, prove that there exists λ C and µ C eigenvalues of A and B respectively, such that α = λ + µ. Solution: This result is well known, here we present an elementary proof. The fact that α C is an eigenvalue of A + B means that the system (A + B) X = αx has a nonzero solution, which we continue to denote with X.

5 We can readily obtain from here that MATH PROBLEMS 5 n 2 A n B = na(na n B) = s n nab n = s n s B n+, n 2 A n B = n 2 s n B n+ On the other hand, we can prove that s s n > n 2 s n, which by above implies that B n+ =. Indeed, to show that s s n > n 2 s n, we first notice that if s n =, the inequality to prove is trivial, as only one of the numbers x k could be zero. On the other hand, if s n >, the inequality to be proved becomes ( (x x n ) ) > n 2, x x n which is true as the numbers are distinct. Hence, B n+ = O, which immediately implies B n = O as well, as the minimal polynomial and the characteristic polynomial of B have the same irreducible factors. As A n = s n B n it follows that A n = O, as well. 5. We denote by M n (R), the set of n n matrices with real entries. Assume a function f : M n (R) R has the properties that f (X) for all X O and that f(xy ) = f(x)f(y ), for any X, Y M n (R). Show that f(x) =, for all X M n (R). Solution: Let us first find f(o). Since f(o) = f(o)f(o), it follows f(o) {, }. Let us assume by contradiction that f(o) =. Let A, B M n (R), such that both A, B O but AB = O. For example, B could have zeros everywhere except the first row, and A could have arbitrary entries except on the first column, where all entries are taken to be zero. Then clearly AB = O, and = f(o) = f(ab) = f(a)f(b), which means that f(x) =, for some nonzero X. This is a contradiction. Hence, f(o) = and now we notice that for all X. = f(o) = f(ox) = f(o)f(x) = f(x), 6. Let A, B, C be n n matrices. Assume that two of the matrices A, B, C commute, where C := AB BA. Prove that C n = O. Solution: For simplicity, we can assume that A commutes with C. We use that for two matrices X, Y the trace tr (XY Y X) =. So, trc =, and for any k >, we

8 8 OVIDIU MUNTEANU. Let (G, ) be a group. We assume that G has an odd number of elements and that there exists a G and n N, such that: a n x = x a, for all x G\{a k k N}. Show that ax = xa, for any x G. Solution: Let us denote with H := {a k k N}. We know that a n = xax, for any x G\H. Since H is a subgroup of G, this implies that if x / H, then x / H as well. From above, this means that we also have a n = x ax, for any x G\H. Therefore, this means x ax = xax, which implies that x 2 a = ax 2, for any x G\H. Now let s fix x G\H, and assume by contradiction that x 2 H. We know that G has an odd number of elements, say that G has = 2p + elements. If x 2 H, it follows that (x 2 ) p H. Since x G\H, we see that x 2p+ = x 2p x G\H. However, x 2p+ = e H, because G has order 2p +. This provides a contradiction, hence the assumption that x 2 H is false. We have showed that x G\H implies x 2 G\H as well. It results that a n = x 2 a(x 2 ). Since x 2 and a commute, this implies a n = a. This immediately implies that ax = xa, for any x G\H, which solves the problem. 2. Let (G, ) be a group with n elements, where n is not a multiple of 3. For a subset H of G we assume that x y 3 H, for any x, y G. Prove that H is a subgroup of G. Solution. Consider an element a H. We then have that a a 3 H, so a 2 H. Continuing this, we get (a 2 ) (a 2 ) 3 H, which means that a 4 H. From a 4 H and a H it follows that (a 4 ) a 3 H, therefore a H. We have thus established that a H, for any a H. Now consider any two x, y H. Since x H, we know that x H as well. Now x H and y H imply that (x ) y 3 H. This proves that xy 3 H, for any x, y H. Now we use induction on p N to prove that xy 3p H, for any x, y H. Indeed, by induction we have that xy 3(p ) H, and since y H, we get that xy 3p = ( xy 3(p )) y 3 H. We discuss two cases: Case I: n = 3k +, for some k N. Taking p = k above we get, for any x, y H xy = xy n H.

9 MATH PROBLEMS 9 Indeed, we have used that y n = e, as G has n elements. This proves H is a subgroup of G. Case II: n = 3k + 2, for some k N. Taking p = k + we get for any x, y H that xy = xy n+ = xy 3k+3 H. Again, this proves that H is a subgroup of G, because we have already established that a H, for any a H. 3. Let G be a group. We assume there exists an homomorphism of groups f : G G G and a G such that f(a, x) = f(x, a) = x, for any x G. Prove that G is abelian. Solution: Recall that G G is a group with respect to Since f is a homomorphism, we have It follows therefore that (x, y) (z, w) = (xz, yw). f(xz, yw) = f(x, y) f(z, w). f(x, e) f(a, a) = f(xa, ea) = f (xa, a) = xa, where e denotes the identity element in G. Moreover, we also know by hypothesis that f(a, a) = a, which by above implies that Similarly, which then implies that In we first take z = e and y = e to get that f(x, e) = x, for any x G. f(e, x) f(a, a) = f(a, xa) = xa, f(x, e) = f(e, x) = x, for any x G. f(xz, yw) = f(x, y) f(z, w) f(x, w) = x w and now take x = e and w = e to get that f(z, y) = y z.

10 OVIDIU MUNTEANU This clearly implies G is abelian and also that f(x, y) = x y, for any x, y G. 4. Let (G, ) be a group with identity e G. We assume that there exists a surjective endomorphism f : G G, such that H := {x G f(x) = e} has the property that if K is a subgroup of G with H K G, then either K = H or K = G. Prove that G Z or G Z p, for p prime. Solution: We first show that H = {e}. Assuming there exists x H\{e}, we can let K := {x G (f f) (x) = e} and L := {x G (f f f) (x) = e}. These are both subgroups of G as f f and f f f are endomorphisms of G. Furthermore, we can show H K L, with strict inclusions. Indeed, as f is onto, there exists x G such that f(x ) = x. It follows that (f f) (x ) = f(x ) = e, so x K. Since f(x ) = x e it follows that x K\H. Similarly, there exists x 2 G such that f(x 2 ) = x, and the same argument as above implies x 2 L\K. By hypothesis, this is a contradiction. The contradiction is to the assumption that there exists x H\ {e}. Therefore, H = {e} and G has no proper subgroups. This implies the conclusion. 5. Assume (G, ) is a group with an odd number of elements and there exist x, y G such that yxy = x. Prove that y = e, where e is the identity in G. Solution: From yxy = x we successively get that y = (yx) x, and y = x(xy). Equalizing the two formulas we get (yx) x = x(xy). This is equivalent to x(xy) = (yx)x, so we have established that x 2 y = yx 2. By induction, we can show that x 2p y = yx 2p. Indeed, p = is already checked. If we assume now that x 2p y = yx 2p, we checked immediately that x 2(p+) y = (x 2p x 2 )y = x 2p (x 2 y) = x 2p (yx 2 ) = yx 2p x 2 = yx 2(p+). Since G has an odd number of elements, we get from here that xy = yx. By hypothesis, we obtain y 2 = e. However, using again that G has an odd number of elements, we get y = e.

11 MATH PROBLEMS 6. Let (G, ) be a group with identity e and a G\{e}. We assume there exists an integer n 2 such that x n+ a = ax, for any x G. Prove that x n2 = e, for any x G. Solution: In x n+ a = ax, we let substitute xa for x to get: (xa) n+ a = axa. Simplifying, we get (xa) n+ = ax. Since ax = x n+ a, it follows that (xa) n+ = x n+ a. We rewrite this as x(ax) n a = xx n a, which simplifying again, implies (ax) n = x n. We have proved that { (ax) n+ = (ax) n ax = x n ax (ax) n+ = ax(ax) n = axx n This implies x n ax = axx n, which gives x n a = ax n. Hence, x n and a commute, for any x G. In x n+ a = ax we substitute x n for x to see that (x n ) n+ a = ax n. Since x n a = ax n, this implies x n(n+) a = x n a. We obtain that x n2 = e, for any x G. 7. Find all groups (G, ) that are the union of three proper subgroups, one of which has a prime number of elements. Solution: We know G = H H 2 H 3, where H 3 = p for a prime number p. Let x H \(H 2 H 3 ) x 2 H 2 \(H H 3 ). For example, such x exists, as otherwise it will imply that H (H 2 H 3 ) so G = H 2 H 3. However, it is known that a group cannot be written as the union of two proper subgroups. The existence of x 2 is established similarly. Now, if x x 2 H, since x H we get x 2 H as well. This is a contradiction to the definition of x 2. Similarly, one can see that x x 2 H 2 is not possible either. Hence, this leaves us with x x 2 H 3 \(H H 2 ). Let us denote a = x x 2. Since H 3 = p and p is prime, it follows that Furthermore, for any k {,..., p }, H 3 = {e, a, a 2,..., a p }. a k H 3 \(H H 2 ). Indeed, if b := a k, then we have H 3 = {e, b,..., b p }. This shows that if b H H 2, then H 3 (H H 2 ), which is a contradiction because G would be the union of H and H 2.

14 4 OVIDIU MUNTEANU Finally, this implies that (xy) 24 = ((xy) 2 ) 67 = (x 2 y 2 ) 67 = (x 2 ) 67 (y 2 ) 67 = x 24 y 24. Now we use that G has 22 elements, so a 24 = a 2, for any a G. By above, this yields that: (xy) 2 = x 2 y 2, for any x, y G. This immediately implies G is abelian. Since G = 22 = 2 9, so G is a free of squares it follows by a know result that G is isomorphic to Z 22.. Let (A, +, ) be a ring with an odd number of elements. We denote by I := {a A a 2 = a}. Prove the following: a) A 3 I 3. b) Determine A provided A = 3 I 3. Solution: a) We know A has an odd number of elements, say A = 2n +. Then we see that 2 (n + ) =, so x = 2 is invertible in A. Let us denote by M := {x A x 2 = }. For any x M, we have ( 2 ( + x) ) 2 = 4 ( + 2x + ) = 2 ( + x), hence 2 ( + x) I. We define the function f : M I by f(x) := 2 ( + x), which is obviously bijective. Hence, M = I. Define also J := {a A a I}. Let us observe that I J = {}; I M = {} and J M = { }. Indeed, if a I J, then a = a = a 2. As 2 is invertible, it results that a =. Now if a I M, we get a 2 = and a 2 = a, so a =. Since I, J, M have the same number of elements, this yields that A 3 I 3. b) If A has exactly three elements, it follows that A is isomorphic to Z 3. Assuming A > 3, we can show that A is isomorphic to Z 3 Z 3. From the proof of a), we know that A = I J M. Therefore, for any x A, we have that x 2 {, x, x}. Since 2 is invertible in A, from here we immediately get that 3 =. Furthermore, we observe that the invertible elements of A must belong to M, so (M, ) forms an abelian group. Choose any a I\ {, }. Since f is bijective, there exists a corresponding y M such that a = y. Now let x M be arbitrary. Since x and y commute, we see that x and a commute. If ax is invertible, it would mean a is invertible as well,

15 MATH PROBLEMS 5 so a =. This is a contradiction, which means ax I J. Let us assume ax I, therefore ax = (ax) 2 = a 2 x 2 = a. Now notice that (a x) 2 = a 2a + = a; and on the other hand, (a x) 2 {, a x, x a}. This implies that x {, a }. In the second case, when ax J, a similar computation gives that x {, a + }. Since a was fixed, it follows that M {,, a, a + }. It is easy to check the converse inclusion is true as well, therefore M = 4, and now A Z 3 Z 3 follows directly from here.. Let (A, +, ) be a commutative ring, having 2n+ elements, where n 4. Let us assume that if a A is not invertible, then a 2 { a, a}. Prove that A is a field. Solution. Let us assume by contradiction that A is not a field. Then there exists a I\ {, }, where I := {a A : a 2 = a}. Indeed, we know there exists x A\{, }, which is not invertible. By hypothesis, it follows that x 2 { x, x}. Then, for either a = x or a = x, we can arrange that a 2 = a. So the set I has at least three elements. Now take any x A, which is invertible. The element ax cannot be invertible, because a is not. Therefore, (ax) 2 {ax, ax}. Using that A is commutative, it follows that ax {a, a}. We now use that the group (A, +) has 2n + elements, which in particular implies that 2 (n + ) =. Hence, this shows that 2 is invertible in A, which from above we know it implies that 2a = a. This means that 3 A is not invertible, which by hypothesis it implies that 3 2 {3, 3}. In both cases we get that 3 =. Recall that for any x A invertible, we have ax {a, a}. If ax = a, it follows that (x a) 2 = x 2 2ax + a 2 = x 2 2a + a 2 = x 2 a 2 = (x a) (x + a). We see that x a cannot be invertible, as otherwise we get x a = x + a, so a =. Hence, since x a is not invertible, we get (x a) 2 {x a, x + a}. Consequently, this means that x 2 a {x a, x+a}. If x 2 a = x a, then x =, as x is invertible. If x 2 a = x + a then x (x + ) = a which implies that x + is not invertible. In this case, we use the hypothesis again to conclude that (x + ) 2 {x +, x }, which now implies that x 2 =. Indeed, if (x + ) 2 = x +, then x 2 + x =, which

16 6 OVIDIU MUNTEANU implies x =. If (x + ) 2 = x, we find that x 2 + 3x + 2 =. Using that 3 = in A, it follows that x 2 =. As we have established above that x 2 + x + a =, using x 2 =, we get that x = a. Concluding, this shows that if ax = a, then x {, a }. On the other hand, if ax = a, then we proceed similarly for (x + a) 2 = x 2 a 2, and get that x {, a + }. This proves that any invertible element of A belongs to {,, a, a + } and the converse is easily checked as well. Certainly, if b I\ {, } is different from a, the argument above shows that thus b = a. Therefore, {,, a, a + } = {,, b, b + }, A = {,,, a, a, a, a +, a, a + } which is a contradiction because A cannot have 9 elements. Thus A is a field. 2. Let (A, +, ) be a ring without zero divisors and a A\ {}. Assume that there exists n N so that x n+ a = ax, for any x A. Prove that A is a field. Solution: In x n+ a = ax, we take x = xa and get: (xa) n+ a = axa. Simplifying by a, as A does not have zero divisors, we get (xa) n+ = ax. Moreover, we have ax = x n+ a, which yields (xa) n+ = x n+ a. We rewrite this as x ((ax) n x n ) a =. Again, simplifying by a and x, we see that (ax) n = x n, for any x A\{}. This proves that { (ax) n+ = (ax) n ax = x n ax which implies that (ax) n+ = ax(ax) n = axx n x n ax = axx n. We rewrite this as (x n a ax n )x =, which implies that x n a = ax n, for any x A\ {}. In x n+ a = ax we now make x = x n to obtain: (x n ) n+ a = ax n. By this, we get x n(n+) a = x n a, which means x n (x n2 )a =. Now we notice that if x A\{}, then x n A\{} as well. We can therefore conclude that This means A is a field. x n2 =, for any x A\{}.

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